SOLUTION: Hi, The question I'm having trouble with is, solve {{{ sin(2x)= sqrt(2)cos(x) }}}, for 0 < x < 2(pi) I've tried expand and collecting to one side; {{{ sin(x)cos(x)+ cos(x)si

Algebra ->  Trigonometry-basics -> SOLUTION: Hi, The question I'm having trouble with is, solve {{{ sin(2x)= sqrt(2)cos(x) }}}, for 0 < x < 2(pi) I've tried expand and collecting to one side; {{{ sin(x)cos(x)+ cos(x)si      Log On


   

Question 635686: Hi,
The question I'm having trouble with is,
solve +sin%282x%29=+sqrt%282%29cos%28x%29+, for 0 < x < 2(pi)
I've tried expand and collecting to one side;
+sin%28x%29cos%28x%29%2B+cos%28x%29sin%28x%29-sqrt%282%29cos%28x%29=+0+
but I've found that that taking cos(x) as a factor will only lead bead to the beginning.
Thankyou
Found 2 solutions by jsmallt9, ewatrrr:
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
+sin%282x%29=+sqrt%282%29cos%28x%29+, for 0 < x < 2(pi)
> I've tried expand and collecting to one side;
+sin%28x%29cos%28x%29%2B+cos%28x%29sin%28x%29-sqrt%282%29cos%28x%29=+0+
Good idea (although you could have left the 2sin(x)cos(x))

> but I've found that that taking cos(x) as a factor will only lead bead to the beginning.
?? Actually factoring out cos(x) is the right thing to do:
+cos%28x%29%28sin%28x%29%2B+sin%28x%29-sqrt%282%29%29=+0+
We now have a product that equals zero. (This is why we make one side of the equation zero.) Then we factor the other side and then use the Zero Product Property:
cos%28x%29+=+0 or %28sin%28x%29%2Bsin%28x%29-sqrt%282%29%29+=+0
Simplifying the second equation:
cos%28x%29+=+0 or 2sin%28x%29-sqrt%282%29+=+0
Solving the second equation for sin(x):
cos%28x%29+=+0 or 2sin%28x%29=sqrt%282%29
cos%28x%29+=+0 or sin%28x%29=sqrt%282%29%2F2

Next we find the general solution. We should recognize that 0 is a special angle value for cos and that sqrt%282%29%2F2 is also a special angle value for sin. So we will not be using our calculators.

For cos(x) = 0:
Only 0 (and angles co-terminal with zero) have a cos of 0. So:
x+=+0+%2B+2pi%2An

For sin%28x%29+=+sqrt%282%29%2F2:
We should know that an angle whose sin is sqrt%282%29%2F2 will have reference
angle of pi%2F4. We should also know that sin is positive in the 1st and 2nd quadrants. Putting these together we get:
x+=+pi%2F4+%2B+2pi%2An (for the 1st quadrant)
x+=+pi-pi%2F4%2B2pi%2An (for the 2nd quadrant)
Simplifying the second equation we get:
x+=+3pi%2F4%2B2pi%2An (for the 2nd quadrant)

All together, our general solution (all the x's that fit your equation) is:
x+=+0+%2B+2pi%2An
x+=+pi%2F4+%2B+2pi%2An
x+=+3pi%2F4%2B2pi%2An

Now the specific solution (the x's in the specified range). For this we replace the n's in the general solution equation with various integers until we are convinced we have found all the x's between 0 and 2pi. (Note: The range excludes 0 and 2pi. Only the numbers between them.)
From x+=+0+%2B+2pi%2An we get no x's that are between 0 and 2pi. When n is 0 or negative we get an x that is too low and when x is 1 or more we get an x that is too big.

From x+=+pi%2F4+%2B+2pi%2An:
When n = 0, x+=+pi%2F4
When n = 1 or more, we get an x that is too big.
When n = -1 or below, we get an x that is too small.

From x+=+3pi%2F4+%2B+2pi%2An:
When n = 0, x+=+3pi%2F4
When n = 1 or more, we get an x that is too big.
When n = -1 or below, we get an x that is too small.

So there are only two solutions between 0 and 2pi: pi%2F4 and 3pi%2F4

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

+sin%282x%29=+sqrt%282%29cos%28x%29+

+2sin%28x%29cos%28x%29=+sqrt%282%29cos%28x%29+

+2sin%28x%29cos%28x%29-+sqrt%282%29cos%28x%29=+0+  |factoring out cos(x) is the correct idea

+cos%28x%29+%282sin%28x%29-+sqrt%282%29%29=+0+

   cos(x) = 0

0r  sin(x) = sqrt%282%29%2F2

Find x below on the (cosx,sinx) unit circle summary